Question

Find the area of square whose 1 pair of opposite vertices are (8,-1)&(4,-4).


Hyy mates plz ans this ..its urgent ...
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Answer 1

Step-by-step explanation:

Let the square be ABCD such B(4,-4) and D(8,-1) and sides be a

BD = √(x2-x1)²+(y2-y1)²

= √(8-4)²+(-1-(-4))²

= √(4)²+(3)²

= √16+9

= √25 = 5 units

By Pythagoras theorem

BD² = a²+a²

5² = 2a²

a = 5/√2

Area of square = a²

= (5/√2)²

= 25/2

12.5 unit²

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Answer 2

Step-by-step explanation:

Let the square be ABCD such B(4,-4) and

D(8,-1) and sides be m

BD = under root of (x2-x1)²+(y2-y1)²... by distance formula..

= under root of (8-4)²+(-1-(-4)² =

= under root of (4)²+(3)?

= 16+9

= under root of 25 = 5 units

By Pythagoras theorem

BD² = m²+m²

52 = 2m²

m = 5root 2

Area of square = m²

= (5root2)2

= 25/2

12.5 unit²..

sry for late answer pickachu..